Factor completely. $9x^2+6x+1=$
Answer: Both $9x^2$ and $1$ are perfect squares, since $9x^2=({3x})^2$ and $1=({1})^2$. Additionally, $6x$ is twice the product of the roots of $9x^2$ and $1$, since $6x=2({3x})({1})$. $9x^2+6x+1 = ({3x})^2+2({3x})({1})+({1})^2$ So we can use the square of a sum pattern to factor: ${a}^2 +2( a)( b)+ {b}^2 =({a}+{b})^2$ In this case, ${a}={3x}$ and ${b}={1}$ : $ ({3x})^2+2({3x})({1})+({1})^2 =({3x}+{1})^2$ In conclusion, $9x^2+6x+1=(3x +1)^2$ Remember that you can always check your factorization by expanding it.